Respuesta :
Answer:
[tex]t=10.62s[/tex]Explanation: The height of the object dropped from the tower can be modeled using the following equation:
[tex]\begin{gathered} y(t)=y_o+v_ot-\frac{1}{2}gt^2\Rightarrow(1) \\ \end{gathered}[/tex]The values of the known quantities are:
[tex]\begin{gathered} y_o=553m \\ v_o=0 \\ g=9.8ms^{-2} \\ \end{gathered}[/tex]Setting equation (1) equal to 0 and plugging in the known values and solving for t gives the following answer:
[tex]\begin{gathered} (553m)+(0)\cdot t-\frac{1}{2}(9.8ms^{-2})\cdot t^2=0 \\ \therefore\Rightarrow \\ \frac{1}{2}(9.8ms^{-2})\cdot t^2=(553m)\Rightarrow t^2=\frac{(553m)\cdot2}{9.8ms^{-2}}=112.86s^2 \\ t=\sqrt[]{(112.86s^2})=10.62s \\ t=10.62s \end{gathered}[/tex]Therefore the time needed for the object to reach the ground is:
[tex]t=10.62s[/tex]